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Let $\space \displaystyle\sum_{n=1}^{\infty} \space \left | \frac{\cos(n\pi)}{n+1}\right |$. Does this series converge or not?

The serie is valid for the natural numbers, so it can be writen as $\space \displaystyle\sum_{n=1}^{\infty} \space \frac{\left |\cos(n\pi)\right |}{n+1}$.

One knows that $\space 0\leq\left |\cos(n\pi)\right |\leq 1$. By using the asymptotic concept, one can say that the leader terms of the top and bottom of the fraction are $\displaystyle \frac{1}{n}$.

And so, the original serie is equivalent to this one $$\space \displaystyle\sum_{n=1}^{\infty} \space \frac{1}{n}$$

that is a $p$ series, where $p\leq 1$ and so diverges, and also the original series. This is correct? Thanks.

Michael Hardy
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2 Answers2

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No. Since the p-series you've considered was only an upper bound for the series, the fact that it diverges to infinity tells you nothing.

john
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    actually, since the original series is $\sum_{n = 1}^{\infty}\left|\cos n\pi\right|/(n + 1)$, the series is actually $\sum_{n = 1}^{\infty}\left|(-1)^n\right|/(n + 1) = \sum_{n = 1}^{\infty}1/(n + 1)$, which diverges because it is just a shifted harmonic series. The reasoning wasn't totally rigorous in getting to the end, but by comparison to $1/n$ the series does indeed diverge. – Stahl May 31 '13 at 16:52
  • Ahh yes. I was concentrating on the method and not the actual question but you are completely correct. – john May 31 '13 at 16:56
  • @john you are right. By using a graph simulator I could see that the sequence corresponding to the p-serie expression is greater than the sequence of the original serie, for all natural numbers. So the direct comparion test doesn't work. Maybe the limit comparation test? I will try.Thanks –  May 31 '13 at 17:06
  • As has been suggested by others, first consider what values $\cos (n\pi)% takes for integer values of $n$. This should simplify the problem greatly. – john May 31 '13 at 17:10
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Since $|\cos(n\pi)| = 1\;\; \forall n,\;$ your series is equivalent to:

$$\sum_{n = 1}^{\infty}\frac{\left|(-1)^n\right|}{(n + 1)} = \sum_{n = 1}^{\infty} \frac {1}{n+1} $$

You're "spot on" with respect to comparing your given series to the harmonic series: $\;\;\displaystyle \sum_{n = 1}^\infty \dfrac 1n,\;$ but since $\;\dfrac 1{n+1} \lt \dfrac 1n,\;$ you can note that

$$\sum_{\color{blue}{\bf n = 1}}^\infty \frac 1{n+1} \quad = \quad \sum_{\color{blue}{\bf n = 2}}\frac 1n$$ and so, save for one term, your series is the harmonic series, and diverges just as does the "strict" harmonic series.

If you want a tighter argument, you can use the limit comparison test of your series and the "strict" harmonic series to prove your series diverges.

$$\text{Put, say}\;\;a_n = \dfrac 1{n+1},\;\;b_n = \dfrac 1n \iff \dfrac{a_n}{b_n} = \dfrac{1/(n+1)}{1/n} = \dfrac n{n+1}$$

$$\text{Then we have}\quad\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac n{n+1} = 1 < \infty$$

Since the harmonic series given by $\sum b_n$ diverges, so too does your series.

amWhy
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  • $\sum_{n = 1}^{\infty} \frac {1}{n+1} = \sum_{n=2}^\infty \frac1{n}$ so the comparison can be done. – marty cohen May 31 '13 at 17:26
  • Indeed, @martycohen I'm just suggesting the the limit comparison test readily reveals and establishes exactly what we, and the OP, already know: that the comparison with the series is valid. – amWhy May 31 '13 at 17:29
  • It is rather senseless to compare the harmonic series to the harmonic series, don't you agree? It is also senseless to write $\sum \frac 1{n+1}\color{red}{<}\sum \frac 1n $ when neither of those sums is defined. – Pedro May 31 '13 at 17:52
  • @amWhy Still, I think one should merely say that $a_n=\frac{1}{n+1}$ is the harmonic series save term, so it diverges. No more, no less. – Pedro May 31 '13 at 17:55
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    Hand wave? There is no hand waving. The terms are the harmonic series shifted by one. That is all. – Pedro May 31 '13 at 18:01